A = \(\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
= \(\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
= \(\sqrt{x-4}+2+l\sqrt{x-4}-2l\)
(+) với \(l\sqrt{x-4}-2l=\sqrt{x-4}-2\) khi \(x\ge8\)
=> A = \(\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
(+) \(l\sqrt{x-4}-2l=2-\sqrt{x-4}\) khi \(4\le x\le8\)
=> A = \(\sqrt{x-4}+2+2-\sqrt{x-4}=4\)
1) Áp dụng bất đẳng thức Cô - si với 4 số \(\frac{5x}{3};\frac{5x}{3};\frac{5x}{3};\frac{1}{x^3}\) dương ta có:
\(B=\frac{5x}{3}+\frac{5x}{3}+\frac{5x}{3}+\frac{1}{x^3}\ge4\sqrt[4]{\frac{5x}{3}.\frac{5x}{3}.\frac{5x}{3}.\frac{1}{x^3}}=4\sqrt[4]{\frac{125}{27}}\)
=> B nhỏ nhất bằng \(4\sqrt[4]{\frac{125}{27}}\) khi \(\frac{5x}{3}=\frac{1}{x^3}\) => x4 = 3/5 => x = \(\sqrt[4]{\frac{3}{5}}\)
2) ĐK : x > 4
\(A=\sqrt{\left(x-4\right)+2\sqrt{x-4}.2+4}+\sqrt{\left(x-4\right)-2\sqrt{x-4}.2+4}\)
\(A=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
\(A=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)
+) Nếu \(\sqrt{x-4}\ge2\) => x - 4 > 4 => x > 8 thì \(A=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
+) Nếu \(\sqrt{x-4}